A rectangular box with a volume of 684 ftcubed is to be constructed with a square base and top. The cost per square foot for the bottom is 20cents, for the top is 10cents, and for the sides is 2.5cents. What dimensions will minimize the cost?

The dimensions of the rectangular box is 29.08 ft×29.08 ft×4.85 ft.

Minimum cost= 26,779.77 cents.

Step-by-step explanation:

Given that a rectangular box with a volume of 684 ft³.

The base and the top of the rectangular box is square in shape.

Let the length and width of the rectangular box be x.

[since the base is square in shape,  length=width]

and the height of the rectangular box be h.

The volume of rectangular box is = Length ×width × height

                                                       =(x²h) ft³

According to the problem,

.....(1)

The area of the base and top of rectangular box is = x² ft²

The surface area of the sides= 2(length+width) height

                                                =2(x+x)h

                                                =4xh ft²

The total cost to construct the rectangular box is

=[(x²×20)+(x²×10)+(4xh×2.5)] cents

=(20x²+10x²+10xh) cents

=(30x²+10xh) cents

Total cost= C(x).

C(x) is in cents.

∴C(x)=30x²+10xh

Putting

Differentiating with respect to x

To find minimum cost, we set C'(x)=0

ft.

Putting the value x in equation (1) we get

  ≈29.08 ft.

The dimensions of the rectangular box is 29.08 ft×29.08 ft×4.85 ft.

Minimum cost C(x)=[30(29.08)²+10(29.08)(4.85)] cents

                                =29,779.77 cents

lol we need to see what the question is asking about

step-by-step explanation:

step-by-step explanation:

hans did not square the 40-unit side; he merely multiplied it by 2.   incorrect.